If $u = - f^{''}( heta)\sin{ heta + f^{'}( heta)\cos heta}$ and $v = f^{''}( heta)\cos{ heta + f^{'}( heta)\sin heta}$, then $f\left\lbrack \left( \frac{du}{d heta} ight)^{2} + \left( \frac{dv}{d heta} ight)^{2} ight brack^{\frac{1}{2}\ }d heta$ is equal to

Question

If $u = - f^{''}(	heta)\sin{	heta + f^{'}(	heta)\cos	heta}$ and $v = f^{''}(	heta)\cos{	heta + f^{'}(	heta)\sin	heta}$, then $f\left\lbrack \left( \frac{du}{d	heta} ight)^{2} + \left( \frac{dv}{d	heta} ight)^{2} ightbrack^{\frac{1}{2}\ }d	heta$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $u = - f^{''}(	heta)\sin{	heta + f^{'}(	heta)\cos	heta}$
and $v = f^{''}(	heta)\cos{	heta + f^{'}(	heta)\sin	heta}$
On differentiating $w.r.t\ 	heta$ respectively, we get
$\frac{du}{d	heta} = - f^{'''}(	heta)\sin{	heta - f^{''}(	heta)\cos{	heta + f^{''}(	heta)\cos{	heta - f^{'}(	heta)\sin	heta}}}$
$= - f^{'''}(	heta)\sin{	heta - f^{'}(	heta)\sin	heta}$
and $\frac{dv}{d	heta} = f^{'''}(	heta)\cos{	heta - f^{''}(	heta)\sin{	heta + f^{''}(	heta)\sin{	heta + f^{'}(	heta)\cos	heta}}}$
$= - f^{'''}(	heta)\cos{	heta + f^{'}(	heta)\cos	heta}$
$	herefore\left( \frac{du}{d	heta} ight)^{2} + \left( \frac{dv}{d	heta} ight)^{2} = \left\lbrack f^{'''}(	heta) ightbrack^{2} + \left\lbrack f^{'}(	heta) ightbrack^{2} + 2f^{'}(	heta)f^{'''}(	heta)$
$= \left\lbrack f^{'''}(	heta) + f^{'}(	heta) ightbrack^{2}$
Now, $\int_{}^{}\left\lbrack \left( \frac{du}{d	heta} ight)^{2} + \left( \frac{dv}{d	heta} ight)^{2} ightbrack^{\frac{1}{2}}d	heta = \int_{}^{}\left\lbrack f^{'''}(	heta) + f^{'}(	heta) ightbrack d	heta$
$= f''(	heta) + f(	heta) + c$

Q. If $u = - f^{''}(\theta)\sin{\theta + f^{'}(\theta)\cos\theta}$ and $v = f^{''}(\theta)\cos{\theta + f^{'}(\theta)\sin\theta}$, then $f\left\lbrack \left( \frac{du}{d\theta} \right)^{2} + \left( \frac{dv}{d\theta} \right)^{2} \right\rbrack^{\frac{1}{2}\ }d\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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